An airplane is flying horizontally with a constant velocity of 160 m/s at an altitude of 5400 m when it drops a package.
How far ahead of the target along the x direction should the plane be when it releases the package?
Calculate how long it takes to hit the ground. Call that time T.
Use H = (1/2) g T^2
Rearrange it to solve for T. H is 5400 m and you should know what g is.
Multiply T by the horizontal flight velocity, 160 m/s.
That will tell you how far the box moves forward along the flight path while it is falling.
5331 m
To determine how far ahead of the target in the x direction the plane should be when it releases the package, we need to consider the horizontal distance the package will travel before reaching the ground.
The velocity of the airplane is given as 160 m/s, and it is flying horizontally. This means that the horizontal component of its velocity remains constant throughout its flight.
When the package is released, it will also inherit the horizontal velocity of the airplane. Therefore, the horizontal distance it covers is solely determined by the time it takes to reach the ground.
To find this time, we can use the equation of motion for vertical motion, which is:
y = y₀ + v₀t + (1/2)at²
Here:
- y is the vertical displacement (altitude) = 5400 m (initial altitude)
- y₀ is the vertical displacement at t = 0 (which is y₀ = 0, as we are measuring the displacement from the ground)
- v₀ is the initial vertical velocity = 0 (the package is dropped, so it has no initial velocity)
- a is the acceleration due to gravity = -9.8 m/s² (negative because it acts downward)
- t is the time taken
Since the package reaches the ground when y = 0, the equation becomes:
0 = 5400 + 0t + (1/2)(-9.8)t²
Rearranging the equation, we get:
4.9t² = 5400
Solving for t, we find:
t² = 5400 / 4.9
t ≈ √(1102.04)
t ≈ 33.2 s (rounded to one decimal place)
Therefore, the time taken for the package to reach the ground is approximately 33.2 seconds.
Now, we can determine the distance covered in the x direction during this time. Since the horizontal velocity of the airplane is 160 m/s and the time is 33.2 seconds, the distance covered is:
d = v₀t
d = 160 m/s × 33.2 s
d ≈ 5312 m
So, the plane should be approximately 5312 meters ahead of the target in the x direction when it releases the package.